Problem 17 - College Algebra and Calculus: An Applied Approach

Question

Describe the interval(s) on which the function is continuous. Explain why the function is continuous on the interval(s). If the function has a discontinuity, identify the conditions of continuity that are not satisfied. \(f(x)=\frac{x}{x^{2}-1}\)

Step-by-Step Solution

Answer

The function \( f(x) = \frac{x}{{x^2 - 1}} \) is continuous on the intervals \( x < -1 \), \( -1 < x < 1 \), and \( x > 1 \). The points of discontinuity are \( x = 1 \) and \( x = -1 \) where the function does not satisfy the condition of continuity because it is not defined there.

Step 1: Identify where the function is undefined

First, set the denominator of the function equal to zero and solve for \( x \). This will tell us where \( f(x) \) is undefined. So we solve: \( x^{2} - 1 = 0 \) which gives \( x = 1 \) and \( x = -1 \)

Step 2: Determine the continuity intervals

The function will be continuous everywhere except where it is not defined, which are the points \( x = -1 \) and \( x = 1 \). So, the solutions are the intervals \( x < -1 \), \( -1 < x < 1 \), and \( x > 1 \)

Step 3: Check the conditions of continuity

The function \( f(x) = \frac{x}{{x^2 - 1}} \) does not satisfy the continuity condition at \( x = 1 \) and \( x = -1 \) because it is not defined there, which means that the points of discontinuity are \( x = 1 \) and \( x = -1 \)

Key Concepts

Discontinuity

Discontinuity in a function refers to the points where the function is not continuous. A function is continuous when you can draw its graph without lifting the pencil from the paper. However, when there is a break in this graph, it's called a discontinuity. For the given function \( f(x) = \frac{x}{{x^2 - 1}} \), let's examine its discontinuity.
The denominator \( x^2 - 1 \) plays a crucial role. By setting it to zero, we find where the function is undefined: \( x = 1 \) and \( x = -1 \). These points cause a division by zero, which results in discontinuities. Thus, the graph will have breaks at these x-values.

• Discontinuity points: \( x = 1 \) and \( x = -1 \).
• Reason: Denominator equal to zero causes breaks in the graph.

Understanding these points helps to know where the function does not exhibit a smooth graph.

Conditions of Continuity

For a function to be continuous at a point, it must meet three main conditions:

• The function must be defined at the point.
• The limit of the function as it approaches the point from both the left and the right must exist.
• The value of the function at the point must equal this limit.

Let's apply these to our function \( f(x) = \frac{x}{{x^2 - 1}} \). At points \( x = 1 \) and \( x = -1 \), the function is undefined, violating the first condition. Thus, it cannot be continuous at these points as both the numerator and denominator become zero, making the function not exist. Therefore, these are points of discontinuity.

Interval Notation

Interval notation is a way to describe sets of numbers between two endpoints and is often used to express where a function is continuous. Intervals can be open (using parentheses) or closed (using brackets), depending on whether the endpoints are included.
The function \( f(x) = \frac{x}{{x^2 - 1}} \) is continuous wherever it is defined, except at the discontinuity points \( x = -1 \) and \( x = 1 \). Using interval notation, we describe the continuous intervals:

• Left of \(-1\): \((-\infty, -1)\)
• Between \(-1\) and \(1\): \((-1, 1)\)
• Right of \(1\): \((1, \infty)\)

Notice the use of parentheses. They're used because \( x = -1 \) and \( x = 1 \) are not included in the intervals. Hence, these intervals show where the function behaves smoothly without any breaks.